As a particle moves along the number line, its position at time $t$ is $s(t)$, its velocity is $v(t)$, and its acceleration is $a(t)= 3t^2$. If $v(0) = 3$ and $s(0) = 1$, what is $s(2)$ ? $s(2)=~$
The antiderivative of $~a(t)~$ is $~v(t)=t^3+C\,$. We know that $~v(0) = 3\,$, so $~C=3\,$. Therefore, $~v(t) = t^3+3\,$. The antiderivative of $~v(t)~$ is $~s(t)=\dfrac14t^4+3t+K\,$. We know that $~s(0) = 1\,$, so $~K=1\,$. Therefore $~s(t)=\dfrac14t^4+3t+1\,$. Then $~s(2)=4+6+1=11\,$.